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Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

When 1-pentyn...

Question

When 1-pentyne $(A)$ is treated with $4 N$ alcoholic $K O H$ at $175^{o} C$ , it is slowly converted into an equilibrium mixture of $1.3$ % of 1-pentyne( $A$ ), $95.2$ % 2-pentyne ( $B$ ) and $3.5$ % of 1,2-pentadiene ( $C$ ). The equilibrium was maintained at $174^{o} C$ . Calculate $Δ G^{o}$ for the following equilibria:
$B ⇌ A; Δ G_{1}^{o} = ?$
$B ⇌ C; Δ G_{2}^{o} = ?$
From the calculated values of $Δ G_{1}^{o}$ and $Δ G_{2}^{o}$ indicate the order of stability of $(A), (B)$ and $(C)$ . Write a reasonable reaction mechanism showing all intermediates leading to $(A), (B)$ and $(C)$ .

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Solution

$t_{e q}$
$(A) = 1.3$ , $(B) = 95.2$ , $(C) = 3.5$
$K_{e q} = \frac{[B] [C]}{[A]} = \frac{95.2 \times 3.5}{1.3} = 256.31$
for $B ⇌ A$
$K_{1} = \frac{[A]}{[B]} = \frac{[C]}{K_{e q}} = 0.013$
$Δ G_{1}^{o} = - 2.303 R T log K_{1} = - 2.303 \times 8.314 \times 298 log 0.013 = 16.178 k J$
for $B ⇌ C$
$K_{2} = \frac{[C]}{[B]} = \frac{K_{e q} [A]}{{[B]}^{2}} = 0.037$
$Δ G_{2}^{o} = - 2.303 R T log K_{2} = - 2.303 \times 8.314 \times 298 log 0.037 = 12.282 k J$
Stability will lie in the order
$B > C > A$

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teq(A)=1.3, (B)=95.2, (C)=3.5Keq=[B][C][A]=95.2×3.51.3=256.31for B⇌AK1=[A][B]=[C]Keq=0.013ΔGo1=−2.303RTlogK1=−2.303×8.314×298log0.013=16.178kJfor B⇌CK2=[C][B]=Keq[A][B]2=0.037ΔGo2=−2.303RTlogK2=−2.303×8.314×298log0.037=12.282kJStability will lie in the orderB>C>A